1 Answer
Solution :-
Let us consider OAB be the equilateral triangle inscribed in parabola y2 = 4ax.
Let AB intersect the x – axis at point C.
Diagrammatic representation of the ellipse is as follows:
_1639122807.bmp)
Now let OC = k
From the equation of the given parabola, we have,
So, y2 = 4ak
y = ±2\( \sqrt{ak}\)
The coordinates of points A and B are (k, \( 2 \sqrt{ak}\)), and (k, \( -2 \sqrt{ak}\))
AB = CA + CB
= \( 2 \sqrt{ak}+ 2 \sqrt{ak}\)
= \( 4\sqrt{ak}\)
Since, OAB is an equilateral triangle, OA2 = AB2.
Then,
k2 + (\( 2 \sqrt{ak}\))2 = (\( 4 \sqrt{ak}\))2
k2 + 4ak = 16ak
k2 = 12ak
k = 12a
Thus, AB = \( 4\sqrt{ak}\)
= \( 4\sqrt{(a×12a)}\)
= \(4\sqrt{12a^2}\)
= \( 4\sqrt{(4a×3a)}\)
= 4(2)\( \sqrt{3a}\)
= \( 8\sqrt{3a}\)
Hence, the side of the equilateral triangle inscribed in parabola
y2 = 4ax is \( 8\sqrt{3a}\).
Answered by Abhisek | 1 year agoRelated Questions
An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.
Find the area of the triangle formed by the lines joining the vertex of the parabola x2 = 12y to the ends of its latus rectum.
A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.
An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.